(I) A 0.20-F capacitor is desired. What area must the plates have if they are to be separated by a 3.2-mm air gap?

The capacitance of a capacitor relies on the area of the plates and separation distance. If the separation distance increases, the capacitance will reduce.

The relation to find capacitance is given by,

\(C = \frac{{{\varepsilon _0}A}}{d}\) … (i)

Here,\({\varepsilon _0}\)is the permittivity of free space, A is the area of the plates and d is the separation of plates.

The separation distance is,\(d = 3.2\;{\rm{mm}}\).

The capacitance is, \(C = 0.20\;{\rm{F}}\).

From equation (i), the area of the plates is given by,

\(A = \frac{{Cd}}{{{\varepsilon _0}}}\)

Substitute the values in the above expression.

\(\begin{aligned}A &= \frac{{0.20\;{\rm{F}} \times \left( {3.2\;{\rm{mm}} \times \frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)}}{{8.85 \times {{10}^{ - 12}}\;{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}}}\\A &= 7.23 \times {10^7}\;{{\rm{m}}^2}\end{aligned}\)

Thus, the area of the plates is \(7.23 \times {10^7}\;{{\rm{m}}^2}\).